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# Let a, b, and c be positive integers with a> b > c such that a^2 - b^2

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Math Expert
Joined: 02 Sep 2009
Posts: 64172
Let a, b, and c be positive integers with a> b > c such that a^2 - b^2  [#permalink]

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18 Apr 2019, 00:01
00:00

Difficulty:

65% (hard)

Question Stats:

18% (02:12) correct 82% (03:29) wrong based on 11 sessions

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Let a, b, and c be positive integers with a> b> c such that $$a^2 - b^2 - c^2 + ab = 2011$$ and $$a^2 + 3b^2 + 3c^2 - 3ab - 2ac - 2bc = -1997$$. What is a?

(A) 249
(B) 250
(C) 251
(D) 252
(E) 253

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Re: Let a, b, and c be positive integers with a> b > c such that a^2 - b^2  [#permalink]

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18 Apr 2019, 00:22
Awaiting explanation by someone:
I was able to add two equations and come up with

$$a^2−b^2−c^2+ab=2011$$
$$a^2+3b^2+3c^2−3ab−2ac−2bc=−1997$$
------------------------------------------------
$$2a^2+2b^2+2c^2−2ab−2ac−2bc =14$$
or
$$a^2+b^2+c^2−ab−ac−bc =7$$

and
$$a^3+b^3+c^3−3abc=(a+b+c)[a^2+b^2+c^2−ab−bc−ac]$$

$$a^3+b^3+c^3−3abc=7(a+b+c)$$

But could not solve further
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Joined: 18 Nov 2018
Posts: 27
Re: Let a, b, and c be positive integers with a> b > c such that a^2 - b^2  [#permalink]

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18 Apr 2019, 03:24
IMO E

Combine two equations and we have
a^2 + b^2 + c^2 - 2ab - 2ac - 2bc = 14
(a-b)^2 + (b-c)^2 + (a-c)^2 = 14

Since a>b>c and all three are integer, (a-c)^2 is 9 while (b-c)^2 and (a-b)^2 can be either 1 and 4.

Two scenarios: either a=c+3 & b=c+2 or a=c+3 & b=c+1

Replace the scenarios into the equation a^2 -b^2 -c^2 +ab = 2011 and we get only 1 integer solution for c which is 250. a=250+3=253.

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Re: Let a, b, and c be positive integers with a> b > c such that a^2 - b^2   [#permalink] 18 Apr 2019, 03:24